If 20^{2-3x^2} = (40sqrt{5})^{3x^2-2},then x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the equation: $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$.Note that $40\sqrt{5} = 20 	imes 2 	imes \sqrt{5} = 20 	imes \sqrt{4} 	imes \sqrt{5}

Vedclass · Accepted Answer

$\pm \sqrt{\frac{2}{3}}$

Vedclass · Answer

(B) Given the equation: $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$.Note that $40\sqrt{5} = 20 	imes 2 	imes \sqrt{5} = 20 	imes \sqrt{4} 	imes \sqrt{5} = 20 	imes \sqrt{20} = 20^{1} 	imes 20^{1/2} = 20^{3/2}$.Substituting this into the equation:$20^{2-3x^2} = (20^{3/2})^{3x^2-2}$$20^{2-3x^2} = 20^{\frac{3}{2}(3x^2-2)}$Since the bases are equal,we equate the exponents:$2-3x^2 = \frac{3}{2}(3x^2-2)$$2-3x^2 = -\frac{3}{2}(2-3x^2)$$(2-3x^2) + \frac{3}{2}(2-3x^2) = 0$$(2-3x^2)(1 + \frac{3}{2}) = 0$$(2-3x^2)(\frac{5}{2}) = 0$Since $\frac{5}{2} 
eq 0$,we must have $2-3x^2 = 0$.$3x^2 = 2$$x^2 = \frac{2}{3}$$x = \pm \sqrt{\frac{2}{3}}$.

If $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$,then $x$ is equal to

Similar Questions

If $5^{x-1} + 5 \cdot (0.2)^{x-2} = 26$,then the value of $x$ can be:

$\frac{1}{\sqrt{11 - 2\sqrt{30}}} - \frac{3}{\sqrt{7 - 2\sqrt{10}}} - \frac{4}{\sqrt{8 + 4\sqrt{3}}} = $

If ${x^{x\sqrt[3]{x}}} = {(x \cdot \sqrt[3]{x})^x}$,then $x =$

${\frac{{[4 + \sqrt{15}]}^{3/2} + {[4 - \sqrt{15}]}^{3/2}}{{[6 + \sqrt{35}]}^{3/2} - {[6 - \sqrt{35}]}^{3/2}}} = $

Which of the following statements are true and which are false? In each case,give a valid reason for your answer.
$t: \sqrt{11}$ is a rational number.

Vedclass Test Series

Exam Paper Generator

Online Exam Module